hdu-3172 Virtual Friends（并查集+容器）

最新推荐文章于 2021-11-05 15:52:10 发布

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本文介绍了一种通过编程计算在线虚拟社交网络中好友关系规模的方法。针对每一对新建立的好友关系，算法能够即时更新并返回包含这两个人在内的社交网络总人数。利用并查集数据结构实现高效查找和合并操作。

Virtual Friends

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7475 Accepted Submission(s): 2138

Problem Description

These days, you can do all sorts of things online. For example, you can use various websites to make virtual friends. For some people, growing their social network (their friends, their friends' friends, their friends' friends' friends, and so on), has become an addictive hobby. Just as some people collect stamps, other people collect virtual friends.

Your task is to observe the interactions on such a website and keep track of the size of each person's network.

Assume that every friendship is mutual. If Fred is Barney's friend, then Barney is also Fred's friend.

Input

Input file contains multiple test cases.
The first line of each case indicates the number of test friendship nest.
each friendship nest begins with a line containing an integer F, the number of friendships formed in this frindship nest, which is no more than 100 000. Each of the following F lines contains the names of two people who have just become friends, separated by a space. A name is a string of 1 to 20 letters (uppercase or lowercase).

Output

Whenever a friendship is formed, print a line containing one integer, the number of people in the social network of the two people who have just become friends.

Sample Input

1
3
Fred Barney
Barney Betty
Betty Wilma

Sample Output

2
3
4

题意：求出每次加入的两个新的关系时朋友链的长度。因为输入不是数字是字符串，所以想到用map去存。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
#define N 100010
int pre[N];
int sum[N];
map<string,int> p;
int finds(int x)
{
    int r=x;
    while(r!=pre[r])
    {
        r=pre[r];
    }
    int k=x,l;
    while(k!=pre[k])
    {
        l=k;
        k=pre[k];
        pre[l]=r;
    }
    return r;
}
void mix(int a,int b)
{
    int x=finds(a),y=finds(b);
    if(x!=y)
    {
        pre[x]=y;
        sum[y]+=sum[x];
    }
    printf("%d\n",sum[y]);
}
int main()
{
    int t,n;
    char a[25],b[25];
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            scanf("%d",&n);
            p.clear();
            int ans=1;
            for(int i=0;i<N;i++)
                {
                    pre[i]=i;
                    sum[i]=1;
                }
            for(int i=0;i<n;i++)
            {
                 scanf("%s %s",a,b);
                 if(!p[a])
                    p[a]=ans++;
                 if(!p[b])
                    p[b]=ans++;
                 mix(p[a],p[b]);
            }
        }
    }
    return 0;
}